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23-Ind-B1 Reliability and Maintainability · December 2019

Question 9 of 9: $2^3$ Factorial Design — Machine Tool Life

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2019 — 17-Ind-B1 Applied Probability & Statistics. Three-hour, closed-book exam; one of two permitted calculators (Sharp or Casio), one 8.5″×11.0″ aid sheet (both sides), statistical tables supplied. Format: three sections — Section A: do 2 of 4 (20 marks); Section B: do 2 of 3 (40 marks); Section C: do 1 of 2 (20 marks) — a 9-question, 80-mark paper as actually structured by the section table and each question's own printed marks (the front page's own summary line states “Exam: 5 Questions. Total marks: 100”, which is internally inconsistent with 20+40+20=80 from its own section breakdown and every question's printed mark value). All nine questions across the three sections are solved below for completeness.

Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — probability density functions and moments (ch. 4), point/interval estimation (ch. 8), two-sample hypothesis testing and the sign test (ch. 9–10, 16), simple linear regression (ch. 11), single-factor ANOVA and multiple comparisons (ch. 13). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3). Montgomery, Design and Analysis of Experiments (9th ed., Wiley) — Bartlett's test, Tukey's HSD, $2^k$ factorial designs (ch. 3, 6).

Question 9 (Section C.2): $2^3$ Factorial Design — Machine Tool Life (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. 8-run $2^3$ design, 2 replicates each ($n=16$ total):

RunABCRep 1Rep 2Total
1−1−1−1323163
2+1−1−1154358
3−1+1−1353469
4+1+1−1354782
5−1−1+1444589
6+1−1+1403777
7−1+1+16050110
8+1+1+1394180

Find. (a) The AB, AC, BC, ABC columns; (b) contrasts/mean effects for A, AB, ABC; (c) SS for A, AB, ABC; (d) full ANOVA at $\alpha=0.05$; (e) which terms are significant; (f) a fitted regression model.

Approach. Interaction columns are the elementwise product of the parent factor columns; each contrast is $\sum(\text{sign})\times(\text{run total})$, each effect SS is $\text{contrast}^2/(n_{\text{rep}}2^k)$, and the pure-replicate error uses the 8 df left over after the 7 orthogonal effect contrasts are removed from the 15 total df.

  1. (a) Completed design matrix (interaction columns = elementwise sign products).
RunABCABACBCABC
1−−−+++−
2+−−−−++
3−+−−+−+
4++−+−−−
5−−++−−+
6+−+−+−−
7−++−−+−
8+++++++
  1. (b) Contrasts and mean effects. Contrast $=\sum(\text{sign}_i\times\text{Total}_i)$; mean effect $=\text{contrast}/(n_{\text{rep}}2^{k-1})=\text{contrast}/8$. $$C_A=-63+58-69+82-89+77-110+80=\boxed{-34}\qquad \text{effect}_A=-34/8=-4.25$$ $$C_{AB}=63-58-69+82+89-77-110+80=\boxed{0}\qquad \text{effect}_{AB}=0/8=0$$ $$C_{ABC}=-63+58+69-82+89-77-110+80=\boxed{-36}\qquad \text{effect}_{ABC}=-36/8=-4.5$$ The remaining contrasts (already given, all reproduced exactly from the raw run totals): $C_B=54$, $C_C=84$, $C_{AC}=-50$, $C_{BC}=-6$ — filling the table's own blanks: $\ast(B)=54$, $\ast\ast(AB)=0$, $\ast\ast\ast(ABC)=-36$.
  2. (c) Sums of squares, $SS=\text{contrast}^2/(n_{\text{rep}}2^k)=\text{contrast}^2/16$. $$SS_A=\frac{(-34)^2}{16}=\boxed{72.25}\qquad SS_{AB}=\frac{0^2}{16}=0\ (\text{given})\qquad SS_{ABC}=\frac{(-36)^2}{16}=\boxed{81.00}\ (\text{matches given})$$ Filling the table's remaining blanks the same way: $\ast(SS_A)=72.25$, $\ast\ast(SS_C)=(84)^2/16=441.00$, $\ast\ast\ast(SS_{AC})=(-50)^2/16=156.25$. Every SS computed from the raw data (including the already-given $SS_B=182.25$, $SS_{BC}=2.25$, $SS_{ABC}=81$, $SST=1457$) matches the exam's own printed values exactly — this dataset is fully self-consistent.
  3. (d) ANOVA table, $\alpha=0.05$. $SST=1457$ (16 raw observations, 15 df); the 7 orthogonal effect contrasts use 7 df, leaving $df_E=8$ for pure replicate error: $$SSE=SST-\sum_{7\text{ effects}}SS=1457-(72.25+182.25+441.0+0+156.25+2.25+81.0)=1457-935.0=\boxed{522.0}$$ $$MSE=\frac{522.0}{8}=65.25,\qquad F_{0.05,1,8}=5.318$$
SourceSSdfMS$F$Significant?
A72.25172.251.107No
B182.251182.252.793No
C441.001441.006.759Yes
AB0.0010.000.000No
AC156.251156.252.395No
BC2.2512.250.034No
ABC81.00181.001.241No
Error522.00865.25——
Total1457.0015———
  1. (e) Significant terms. Only the main effect $\boxed{C\text{ (cutting angle)}}$ clears $F_{0.05,1,8}=5.318$ ($F=6.76$); every other main effect and interaction (A, B, AB, AC, BC, ABC) is not statistically significant at the 5% level.
  2. (f) Fitted regression model (coded $\pm1$ units). Regression coefficient $=\text{effect}/2$; intercept $=$ grand mean $=39.25$: $$\hat y=39.25-2.125A+3.375B+5.25C+0\cdot AB-3.125\,AC-0.375\,BC-2.25\,ABC$$ Per part (e), only the $C$ term is individually justified by the significance testing; a parsimonious reduced model retaining just the significant term is $\hat y\approx39.25+5.25C$, though the numerically larger AC and B coefficients ($-3.125$ and $3.375$) are worth flagging as practically non-negligible even though this small replicate count ($n_{\text{rep}}=2$) does not give them enough power to clear significance individually.
Final Results
QuantityValue
(b) $\ast,\ast\ast,\ast\ast\ast$ (contrasts)$B=54,\ AB=0,\ ABC=-36$
(c) $\ast,\ast\ast,\ast\ast\ast$ (SS)$SS_A=72.25,\ SS_C=441.00,\ SS_{AC}=156.25$
(d) Error term$SSE=522.00$, $df=8$, $MSE=65.25$
(e) Significant term(s)C only ($F=6.76\gt5.318$)
(f) Fitted model$\hat y=39.25-2.125A+3.375B+5.25C-3.125AC-0.375BC-2.25ABC$
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